est: Chapter 2 Post-lest This Question: 1 pt 21 of 26 (20 complete) This Test: 2
ID: 3069198 • Letter: E
Question
est: Chapter 2 Post-lest This Question: 1 pt 21 of 26 (20 complete) This Test: 26 pts possib and class Full data set boundares for the given frequency distribution Frequency 63 0-679 680-729 73.0-779 78 0-82.9 830-879 88 0-92 9 930-979 98 0-102.9 103 0-1079 108 0-112.9 What is the class width? What are the class midpoints? (Use ascending order. Round to two decimal places as needed ) What are the class boundaries? (Use ascending order Round to two decimal places as needed ) Enter your answer in each of the answer boxes.Explanation / Answer
Gap = (Lower Limit of (n+1)th class - Upper Limit of nth class)
Add Gap/2 to upper limits and subtract Gap/2 from lower limits
(b) and (c) are obtained from the table
For (a), Range = 112.9-63 = 49.9
No. of classes = 10
Class width = 49.9/10 = 4.99
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Class Freq Mid Points Lower Limits Upper Limits Lower Boundaries Upper Boundaries 63-67.9 4 65.45 63 67.9 62.95 67.95 68-72.9 25 70.45 68 72.9 67.95 72.95 73-77.9 1 75.45 73 77.9 72.95 77.95 78-82.9 9 80.45 78 82.9 77.95 82.95 83-87.9 0 85.45 83 87.9 82.95 87.95 88-92.9 0 90.45 88 92.9 87.95 92.95 93-97.9 0 95.45 93 97.9 92.95 97.95 98-102.9 0 100.45 98 102.9 97.95 102.95 103-107.9 0 105.45 103 107.9 102.95 107.95 108-112.9 1 110.45 108 112.9 107.95 112.95 Gap 0.1 Gap/2 0.05Related Questions
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